Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > edge(100100);
vector<int> dp[100100][17 + 1];
vector<int>::iterator it;
int level[100100];
int up[100100][17 + 1];
int u, v, a, n, m, q, i;
void crop(vector<int>& a, int size) {
  while ((int)a.size() > size) {
    a.pop_back();
  }
}
void combine(vector<int>& c, vector<int> a, vector<int> b) {
  c.resize(a.size() + b.size());
  merge((a).begin(), (a).end(), (b).begin(), (b).end(), c.begin());
  c.resize(unique((c).begin(), (c).end()) - c.begin());
  crop(c, 10);
}
void dfs(int u, int p = 1) {
  level[u] = level[p] + 1;
  up[u][0] = p;
  sort((dp[u][0]).begin(), (dp[u][0]).end());
  crop(dp[u][0], 10);
  for (int i = 1; i <= 17; i++) {
    up[u][i] = up[up[u][i - 1]][i - 1];
    combine(dp[u][i], dp[up[u][i - 1]][i - 1], dp[u][i - 1]);
  }
  for (int i = 0; i < edge[u].size(); i++) {
    int to = edge[u][i];
    if (to != p) {
      dfs(to, u);
    }
  }
}
int get_lca(int u, int v) {
  if (level[u] > level[v]) swap(u, v);
  int dif = level[v] - level[u];
  for (int i = 17; i >= 0; i--)
    if (dif >= (1 << i)) {
      v = up[v][i];
      dif -= (1 << i);
    }
  if (u == v)
    return u;
  else {
    for (int i = 17; i >= 0; i--)
      if (up[u][i] != up[v][i]) {
        u = up[u][i];
        v = up[v][i];
      }
    return up[u][0];
  }
}
void add(vector<int>& result, int u, int lca) {
  int dif = level[u] - level[lca];
  for (int i = 17; i >= 0; i--)
    if (dif >= (1 << i)) {
      combine(result, dp[u][i], result);
      u = up[u][i];
      dif -= (1 << i);
    }
  combine(result, dp[u][0], result);
}
void query(int u, int v, int a) {
  int lca = get_lca(u, v);
  vector<int> result;
  add(result, u, lca);
  add(result, v, lca);
  crop(result, a);
  printf("%d ", result.size());
  for (int i = 0; i < result.size(); i++) printf("%d ", result[i]);
  printf("\n");
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  for (i = 1; i <= m; i++) {
    scanf("%d", &u);
    dp[u][0].push_back(i);
  }
  dfs(1);
  for (i = 1; i <= q; i++) {
    scanf("%d%d%d", &u, &v, &a);
    query(u, v, a);
  }
  return 0;
}